Why is the group $GL(n, \mathbb{R})$ of dimension $n^{2}$?
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2$\begingroup$ The group $M(n, \mathbb{R})$ of $n\times n$ matrices has dimension $n^2$. Why? It's easy to show that $M(n, \mathbb{R})$ a vector space. Any element $\in M(n, \mathbb{R})$ can be written in terms of a sum of $n^2$ matrices $B_{ij}$ where the entry $i,j)$ of $B_{ij} = 1$, and zero otherwise. You can show that the set $\{B_{ij}\}$ is a basis of $M(n, \mathbb{R})$, and hence $M(n, \mathbb{R})$ has dimension $n^2$. $GL(n,\mathbb{R})$ is a subset of $M(n, \mathbb{R})$ under the determinant map. It has the same dimension by Steve's answer below. $\endgroup$– user2468Mar 6, 2012 at 21:04
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3 Answers
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It is an open submanifold of the set of $n$ by $n$ matrices, which is a vector space of dimension $n^2$.
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$\begingroup$ Is there are more elementary way to see this though? $\endgroup$– 198203Mar 6, 2012 at 20:47
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7$\begingroup$ @198203: Isn't that plenty elementary already? It's about the most elementary imaginable argument that can reach the relevant definition of "dimension" at all. $\endgroup$ Mar 6, 2012 at 20:54
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1$\begingroup$ How many basis elements does an $n$ by $n$ matrix have? Certainly it must be $n^2$... take the obvious basis: a single entry in each matrix to be $1$ and all others to be $0$. Do this $n \cdot n$ times, one for each position in the matrix. $\endgroup$– TylerMar 6, 2012 at 20:55
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2$\begingroup$ Basically, the determinant is a continuous function from the $n\times n$ matrices to $\mathbb R$, and the invertible ones are the ones with non-zero determinant, so $GL(n)=det^{-1}(\mathbb {R}\setminus 0)$. Since $\mathbb R\setminus 0$ is an open subset of $\mathbb R$, $GL(n)$ has to be an open subset of the set of all matrices. $\endgroup$ Mar 6, 2012 at 22:24
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2$\begingroup$ @198203 Do you know the definition of the dimension of a manifold? Once you have the definition, the fact that an open submanifold of an $n$-manifold has dimension $n$ is essentially immediate. What's non-trivial and requires work is the proof that the notion of the dimension of a manifold is well-defined. $\endgroup$ Mar 6, 2012 at 22:53
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You should notice that the determinant is a continuous map $f:M(n,\mathbb{R})\equiv\mathbb{R}^{n^2}\to \mathbb{R}$, $f(X)=\det(X)$
Then note that
$$GL(n, \mathbb{R})=f^{-1}(\mathbb{R}\setminus\{0\})$$
How the pre-image by continuous map of open set is an open set then $GL(n, \mathbb{R})$ is an open set of $\mathbb{R}^{n^2}$ whose dimension is $n^2$.
(I am here using that an open set of $\mathbb{R}^{k}$ has $k$-dimension.)
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$\begingroup$ How is, by the way, the dimension of an open set defined ? I am wondering because i am familiar with dimensions of vectorspaces etc. but not this. $\endgroup$– user42761Nov 19, 2012 at 6:43
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1$\begingroup$ If a manifold as a topological space is locally homeomorphic to $R^n$ then its dimension is $n$. We can also think about the Haussdorf Dimension. $\endgroup$ Nov 23, 2012 at 18:04
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$\begingroup$ Why does the open set of $R^{n^2}$ has the dimension $R^{n^2}$? If n=2, $\{(x,y,0,0):x,y \in \Bbb R\} $ has basis $(1,0,0,0)$ and $(0,1,0,0)$ . $\endgroup$ Jan 22, 2020 at 16:09
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I'm sorry people misunderstood your question. You are clearly asking about dimension in terms of linear algebra and not in terms of manifolds.
The set $GL(n)$ is not a subspace. To see this, take any invertible matrix $A$, then $-A$ is also invertible. But $A-A=0$ is obviously not invertible, thus $GL(n)$ is not a vector subspace for any $n$ so it makes no sense to talk about it's dimension in terms of independent vectors.
However, GL(n) is what is known as a submanifold which means that although $GL(n)$ is not a vector space, you can "locally" (i.e. around some neighborhood around the set) view it a vector space structure. But this is something you shouldn't worry about in linear algebra.
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1$\begingroup$ I doubt the OP is still wondering this after ~10 years later lol $\endgroup$ Jan 19 at 2:57